[LeetCode 267] Palindrome Permutation II

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

For example:

Given s = "aabb", return ["abba", "baab"].

Given s = "abc", return [].

Hint:

1. If a palindromic permutation exists, we just need to generate the first half of the string.
2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation

Solution:

odd time char must be in the middle position, similar with Palindrome Permutation Question, can use int array to store number of char appear time.

public List<String> generatePalindromes(String s) {        int[] map = new int[256];        int min = Integer.MAX_VALUE;        int max = Integer.MIN_VALUE;        for(char c: s.toCharArray()) {            map[c]++;            min = Math.min(min, c);            max = Math.max(max, c);        }        int count = 0;        List<String> res = new ArrayList<>();        int oddIndex = 0;        for(int i=min;i<=max;i++) {            if(count ==0 && map[i]%2==1) {                oddIndex = i;                count++;            }else if(map[i]%2 == 1){                return res;            }        }        String cur = "";        if(count==1) {            cur += (char)oddIndex;            map[oddIndex]--;        }        dfs(map, cur, s, res);        return res;            }    private void dfs(int[] map, String cur, String s, List<String> res) {        if(cur.length()==s.length()) {            res.add(cur);            return;        }        for(int i=0;i<map.length;i++) {            if(map[i]>0) {                map[i]-=2;                cur = (char)i + cur + (char)i;                dfs(map, cur, s, res);                cur = cur.substring(1, cur.length()-1);                map[i]+=2;            }        }    }